Appendix

From Section 4 we have ${\bf h}_k = \alpha {\bf h}_{k-1} + {\bf v}_k$, where ${\bf v}_k$ is Gaussian with mean $\bf d$ and covariance $\bf C$. Therefore $f({\bf h}_k\vert{\bf h}_{k-1})$ is Gaussian with mean ${\bf d} + \alpha {\bf h}_{k-1}$ and covariance $\bf C$. To derive the formulas for ${\bf g}_k$ and ${\bf G}_k$ in this case, we start by examining the forms of $f({\bf h}_k)$and $f({\bf h}_{k-1}\vert{\bf h}_k)$.

Recursively writing ${\bf h}_k$ in terms of ${\bf h}_{k-1}$, ${\bf h}_{k-2}, \ \ldots$, we obtain:

$${\bf h}_k = \sum_{j=0}^{\infty} \alpha ^j {\bf v}_{k-j} .$$ (20)

Assuming that $\vert\alpha\vert < 1$, the result of the summation in (20) is that $f({\bf h}_k)$ is Gaussian with mean $\frac{\bf d}{1 - \alpha}$and covariance $\frac {\bf C}{1 - \vert\alpha\vert^2}$.

Using Bayes rule, $f({\bf h}_{k-1}\vert{\bf h}_k) = \frac {f({\bf h}_k\vert{\bf h}_{k-1}) f({\bf h}_{k-1})} {f({\bf h}_k)} \doteq e^{- E_{k-1} }$ , with:

$$E_{k-1} = \left( {\bf h}_{k} - {\bf d} - \alpha {\bf h}_{k-1} \ri... ... ^{*T} {\bf C}^{-1} \left( {\bf h}_{k} - {\bf d} - \alpha {\bf h}_{k-1} \right)$$

$$\ + \left( {\bf h}_{k-1} - \frac{\bf d}{1 - \alpha} \right) ^{*T}... ...a\vert^2 ) {\bf C}^{-1} \left( {\bf h}_{k-1} - \frac{\bf d}{1 - \alpha} \right)$$

$$\ - \left( {\bf h}_{k} - \frac{\bf d}{1 - \alpha} \right) ^{*T} (... ...a\vert^2 ) {\bf C}^{-1} \left( {\bf h}_{k} - \frac{\bf d}{1 - \alpha} \right) .$$

Algebraic manipulation yields:

$$E_{k-1} = \left( {\bf h}_{k-1} - \beta {\bf d} - {\alpha}^* {\bf ... ...C}^{-1} \left( {\bf h}_{k-1} - \beta {\bf d} - {\alpha}^* {\bf h}_{k} \right) ,$$

with $\beta \stackrel{\triangle}{=} \frac {1 - \alpha^*} {1 - \alpha}$ . So $f({\bf h}_{k-1}\vert{\bf h}_k)$ is Gaussian with mean $\beta {\bf d} + \alpha^* {\bf h}_k$ and covariance $\bf C$.

Note that $f({\bf H})$ can be written in terms of $f({\bf h}_k)$ for any value of k. (16) shows this for k=1. In general:

 

$$f({\bf H}) f({\bf h}_N\vert{\bf h}_{N-1}) \cdots f({\bf h}_{k+1}\vert{\bf h}_k) f({\bf h}_k) f({\bf h}_{k-1}\vert{\bf h}_k)$$ =

$$\cdot ~ f({\bf h}_{k-2}\vert{\bf h}_{k-1}) \cdots f({\bf h}_1\vert{\bf h}_2)$$ (21)

To evaluate $f({\bf h}_k \vert {\bf r,B})$, we can use the above form of $f({\bf H})$ in (16) and integrate over ${\bf h}_i, i=1,\ldots,N; i \ne k$.

Consider two of these integrations, say the integrals over ${\bf h}_1$ and ${\bf h}_2$. By examining the form of this result, we will be able to produce the general result. Showing just the terms involving ${\bf h}_1$, $I_1 = \int_{{\bf h}_1} e^{- E_1 } d{\bf h}_1$ , with:

$$({\bf h}_1 - \beta {\bf d} - \alpha^* {\bf h}_2)^{*T} {\bf C}^{-1} ({\bf h}_1 - \beta {\bf d} - \alpha^* {\bf h}_2)$$ E₁ =

$$+ \frac {\vert r_1 - {\bf b}_1^T {\bf h}_1\vert^2 } {\sigma^2} .$$

E₁ can be reduced, dropping constant terms, to:

$$\doteq ({\bf h}_1 - {\bf\hat{g}}_1)^{*T} {\bf\hat{G}}_1^{-1} ({\bf h}_1 - {\bf\hat{g}}_1)$$ E₁

$$- {\bf\hat{g}}_1^{*T} {\bf\hat{G}}_1^{-1} {\bf\hat{g}}_1 + \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf h}_2$$

$$+ \alpha \beta {\bf h}_2^{*T} {\bf C}^{-1} {\bf d} + \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_2 ,$$

where ${\bf\hat{G}}_1$ and ${\bf\hat{g}}_1$ will be defined below. Since the last four terms in E₁ are not functions of ${\bf h}_1$, they can be factored out of the integral. And the integral over ${\bf h}_1$ involving the first term from E₁produces a constant which can be ignored. Thus, the result is:

 

$$\doteq e^{ {\bf\hat{g}}_1^{*T} {\bf\hat{G}}_1^{-1} {\bf\hat{g}}_1 - \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf h}_2 }$$ I₁

$$\cdot ~ e^{ - \alpha \beta {\bf h}_2^{*T} {\bf C}^{-1} {\bf d} - \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_2 } ,$$ (22)

with:

 

$${\bf\hat{G}}_1 \left[ \frac {{\bf b}_1^* {\bf b}_1^T} {\sigma^2} + {\bf C}^{-1} \right]^{-1}$$ =

$${\bf\hat{q}}_1 \frac {r_1 {\bf b}_1^* } {\sigma^2}+ \beta {\bf C}^{-1} {\bf d}$$ = (23)

$${\bf\hat{g}}_1 {\bf\hat{G}}_1 ( {\bf\hat{q}}_1 + \alpha^* {\bf C}^{-1} {\bf h}_2 )$$ =

Expanding the first term from the exponent in (22), and dropping constant terms:

$${\bf\hat{g}}_1^{*T} {\bf\hat{G}}_1^{-1} {\bf\hat{g}}_1 \left( {\bf\hat{q}}_1^{*T} + \alpha {\bf h}_2^{*T} {\bf C}^{-1} \... ... {\bf\hat{G}}_1 \left( {\bf\hat{q}}_1 + \alpha^* {\bf C}^{-1} {\bf h}_2 \right)$$ =

$$\doteq \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1 {\bf C}^{-1} {\bf h}_2$$

$$+ \alpha^* {\bf\hat{q}}_1^{*T} {\bf\hat{G}}_1 {\bf C}^{-1} {\bf h}_2 + \alpha {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1 {\bf\hat{q}}_1 .$$

So we now have:

$$\doteq e^{ \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1... ...\bf h}_2 + \alpha^* {\bf\hat{q}}_1^{*T} {\bf\hat{G}}_1 {\bf C}^{-1} {\bf h}_2 }$$ I₁

$$\cdot ~ e^{ \alpha {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1 {\bf\hat{q}}_1 - \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf h}_2 }$$

$$\cdot ~ e^{ - \alpha \beta {\bf h}_2^{*T} {\bf C}^{-1} {\bf d} - \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_2 }$$

This result shows how the integral over ${\bf h}_1$ produces terms involving ${\bf h}_2$ which must be included when performing the integral over ${\bf h}_2$. Showing just the terms involving ${\bf h}_2$ for this next integral, $I_2 = \int_{{\bf h}_2} e^{- E_2 } d{\bf h}_2$ , with:

$$({\bf h}_2 - \beta {\bf d} - \alpha^* {\bf h}_3)^{*T} {\bf C}^{-1} ({\bf h}_2 - \beta {\bf d} - \alpha^* {\bf h}_3)$$ E₂ =

$$+ \frac {\vert r_2 - {\bf b}_2^T {\bf h}_2\vert^2 } {\sigma^2} - ... ...\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1 {\bf C}^{-1} {\bf h}_2$$

$$- \alpha^* {\bf\hat{q}}_1^{*T} {\bf\hat{G}}_1 {\bf C}^{-1} {\bf h}_2 - \alpha {\bf h}_2^{*T} {\bf C}^{-1} {\bf\hat{G}}_1 {\bf\hat{q}}_1$$

$$+ \vert\alpha\vert^2 {\bf h}_2^{*T} {\bf C}^{-1} {\bf h}_2 + \alpha \beta {\bf h}_2^{*T} {\bf C}^{-1} {\bf d}$$

$$+ \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_2$$

E₂ can be reduced, dropping constant terms, to:

 

$$\doteq ({\bf h}_2 - {\bf\hat{g}}_2)^{*T} {\bf\hat{G}}_2^{-1} ({\bf h}_2 - {\bf\hat{g}}_2) - {\bf\hat{g}}_2^{*T} {\bf\hat{G}}_2^{-1} {\bf\hat{g}}_2$$ E₂

$$+ \vert\alpha\vert^2 {\bf h}_3^{*T} {\bf C}^{-1} {\bf h}_3 + \alpha \beta {\bf h}_3^{*T} {\bf C}^{-1} {\bf d}$$

$$+ \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_3 ,$$ (24)

where ${\bf\hat{G}}_2$ and ${\bf\hat{g}}_2$ will be defined below. Since the last four terms in (24) are not functions of ${\bf h}_2$, they can be factored out of the integral. And the integral over ${\bf h}_2$involving the first term produces a constant which can be ignored. Thus, the result is:

 

$$\doteq e^{ {\bf\hat{g}}_2^{*T} {\bf\hat{G}}_2^{-1} {\bf\hat{g}}_2 - \vert\alpha\vert^2 {\bf h}_3^{*T} {\bf C}^{-1} {\bf h}_3 }$$ I₂

$$\cdot ~ e^{ - \alpha \beta {\bf h}_3^{*T} {\bf C}^{-1} {\bf d} - \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_3 }$$ (25)

with:

 

$${\bf\hat{G}}_2 \left[ \frac {{\bf b}_2^* {\bf b}_2^T} {\sigma^2} + {\bf C}^{-1} ... ...t( {\bf C}^{-1} - {\bf C}^{-1} {\bf\hat{G}}_1 {\bf C}^{-1} \right) \right]^{-1}$$ =

$${\bf\hat{q}}_2 \frac {r_2 {\bf b}_2^* } {\sigma^2}+ (1 - \alpha^*) {\bf C}^{-1} {\bf d} + \alpha {\bf C}^{-1} {\bf\hat{G}}_1 {\bf\hat{q}}_1$$ = (26)

$${\bf\hat{g}}_2 {\bf\hat{G}}_2 ( {\bf\hat{q}}_2 + \alpha^* {\bf C}^{-1} {\bf h}_3 )$$ =

From the forms of (22) and (25) we can easily deduce the general form of the result of integrating over ${\bf h}_i$.

(26) indicates the general form of the equations for ${\bf\hat{G}}_i$ and ${\bf\hat{g}}_i$. These equations have more terms than (23) because in deriving (23) there were no previous integration results to incorporate. However, if we define ${\bf\hat{G}}_0 = {\bf C}$ and ${\bf\hat{q}}_0 = \beta {\bf C}^{-1} {\bf d}$, then (23) may also be written in the same form as (26).

The preceding derivations used (21) to integrate (16) over ${\bf h}_1$ and ${\bf h}_2$, and show the recursive forms of the update equations which appear in (19). If we start at the ``other end'' of (21), and integrate over ${\bf h}_N$ first, then ${\bf h}_{N-1}$, etc., we obtain similar recursive equations which appear in (17).

Finally, to produce $f({\bf h}_k\vert{\bf r},{\bf B})$after integrating over ${\bf h}_i, i = 1,\ldots,N, i \ne k$, we have $f({\bf h}_k\vert{\bf r},{\bf B}) \doteq e^{ {-E_k}}$, with:

$$\left( {\bf h}_{k} - \frac{\bf d}{1 - \alpha} \right) ^{*T} \left... ...rt^2 \right) {\bf C}^{-1} \left( {\bf h}_{k} - \frac{\bf d}{1 - \alpha} \right)$$ E_k =

$$+ \frac {\vert r_k - {\bf b}_k^T {\bf h}_k\vert^2 } {\sigma^2}$$

$$+ \vert\alpha\vert^2 {\bf h}_k^{*T} {\bf C}^{-1} {\bf h}_k + \alp... ...\bf h}_k^{*T} {\bf C}^{-1} {\bf d} + \alpha {\bf d}^{*T} {\bf C}^{-1} {\bf h}_k$$

$$- \alpha {\bf q}_{k+1}^{*T} {\bf\hat{G}}_{k+1} {\bf C}^{-1} {\bf h}_k - \alpha^* {\bf h}_k^{*T} {\bf C}^{-1} {\bf\hat{G}}_{k+1} {\bf q}_{k+1}$$

$$- \vert\alpha\vert^2 {\bf h}_k^{*T} {\bf C}^{-1} {\bf\hat{G}}_{k+1} {\bf C}^{-1} {\bf h}_k$$

$$+ \vert\alpha\vert^2 {\bf h}_k^{*T} {\bf C}^{-1} {\bf h}_k + \alpha \beta {\bf h}_k^{*T} {\bf C}^{-1} {\bf d}$$

$$+ \alpha^* \beta^* {\bf d}^{*T} {\bf C}^{-1} {\bf h}_k - \alpha^* {\bf q}_{k-1}^{*T} {\bf\hat{G}}_{k-1} {\bf C}^{-1} {\bf h}_k$$

$$- \alpha {\bf h}_k^{*T} {\bf C}^{-1} {\bf\hat{G}}_{k-1} {\bf q}_{k-1}$$

$$- \vert\alpha\vert^2 {\bf h}_k^{*T} {\bf C}^{-1} {\bf\hat{G}}_{k-1} {\bf C}^{-1} {\bf h}_k$$

where the third through fifth lines above show the cumulative contributions of the integrals over ${\bf h}_i$ for i>k, and the last four lines show the contributions from i<k. Combining terms and dropping constants, this can be written as:

$$E_k \doteq ({\bf h}_k - {\bf g}_k)^{*T}{\bf G}_k ({\bf h}_k - {\bf g}_k)$$

indicating that $f({\bf h}_k\vert{\bf r},{\bf B})$ has a Gaussian distribution, with ${\bf G}_k$ and ${\bf g}_k$ defined in (18).